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CCGPS Coordinate Algebra Day 2 (8-14-12)

CCGPS Coordinate Algebra Day 2 (8-14-12). UNIT QUESTION: Why is it important to understand the relationship between quantities? Standard: MCC9-12.N.Q.1-3, MCC9-12.A.SSE.1, MCC9-12.A.CED.1-4 Today’s Question: How are unit conversions performed, and why is it important?

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CCGPS Coordinate Algebra Day 2 (8-14-12)

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  1. CCGPS Coordinate AlgebraDay 2 (8-14-12) UNIT QUESTION: Why is it important to understand the relationship between quantities? Standard: MCC9-12.N.Q.1-3, MCC9-12.A.SSE.1, MCC9-12.A.CED.1-4 Today’s Question: How are unit conversions performed, and why is it important? Standard: MCC9-12.N.Q.1 and N.Q.2

  2. Measurements Problem Solving Using Conversion Factors

  3. Example 1 1. Bob studied for 2.5 hrs. How many minutes did he study for? Multiply by: What you want What you have

  4. How many minutes are in 2.5 hours? Initial unit 2.5 hr Conversion Final factor unit 2.5 hr x 60 min = 150 min 1 hr cancel

  5. Learning Check A rattlesnake is 2.44 m long. How long is the snake in cm? A) 2440 cm B) 244 cm C) 24.4 cm

  6. Solution A rattlesnake is 2.44 m long. How long is the snake in cm? B) 244 cm 2.44 m x 100 cm = 244 cm 1 m

  7. Example 2 How many seconds are in 1.4 days? Unit plan: days hr min seconds LecturePLUS Timberlake

  8. Solution Unit plan: days hr min seconds 1.4 day x 24 hr x 60 min x 60 sec 1 day 1 hr 1 min = 120,960sec

  9. Learning Check If the ski pole is 3.0 feet in length, how long is the ski pole in mm?

  10. Solution 3.0 ft x 12 in x 2.54 cm x 10 mm = 1 ft 1 in. 1 cm = 914.4 mm

  11. Example 3 John Isner serves 140 miles per hour. How fast is that feet per second?

  12. Solution 140 miles x 5,280 ft. x 1 hr x 1 min = 1 hr 1 mile 60 min 60 sec. = 205.3 ft/sec.

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